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3 years ago

What is the acceleration of a 1107 kg car as it comes to rest with a force of 2242 N in 6.9 seconds?

Physics

1 answer:

xz_007 [3.2K]3 years ago

4 0

acceleration is 2.025m/s².

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A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass = 5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate

and

here work need to lift cable is

w = (10Δy )(9.8 y ) j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= $\int\limits^{10}_0 {98y} \, dy$

= $[49 y^2]^{50}_0$

= 2450J

so lifting 5000 lb wrcking 50 m required additional 5000 + 2450

so total work is = 52450 J

3 0

3 years ago

What happens when you increase energy efficiency? (Select the best answer.)

IRISSAK [1]

You use less energy to provide the same service.

3 0

1 year ago

Need help science the top and bottom! Please

Oksi-84 [34.3K]

errrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

7 0

3 years ago

You have DNA samples from five different suspects. You know that all humans have very similar DNA. What feature of DNA allows yo

Aneli [31]

Answer:

The answer is positively B.

There are so many base pairs that there are individual differences.

Confirmed by my Forensics test today.

Explanation:

4 0

3 years ago

Light of wavelength 478 nm in air falls on two slits 6.53 mm apart. The slits are immersed in water, as is a viewing screen 30.8

cluponka [151]

The distance between the two fringes is 0.0225 mm.

<u>Explanation:</u>

As per Young's double slit experiment, the distance between the fringes can be observed for the bright constructive interference fringes. So if we know the distance between the separation of slits. the distance of the mirror placed away from the slit, then the distance between two fringes can be observed using the below formula.

$y=\frac{m\times wavelength\times D}{d}$

Here y is the distance between the fringes on screen, D is the distance of the screen from the slits and d is the distance between the slits. Also m is the order of the interference which is generally considered as 1.

So, in the present case, D = 30.8 cm = 0.308 m, d = 6.53×10⁻³ m, wavelength = 478 nm and m = 1.

So, $y = \frac{1\times 478\times 10^{-9} \times 0.308}{6.53\times 10^{-3} }$

$y =22.546\times 10^{-9+3}=22.546\times 10^{-6} m=0.0225\ mm.$

So, the distance between the two fringes is 0.0225 mm.

6 0

3 years ago